y=ln根号下1+sinx/1-sinx的微分

y=1/2 ln(1+sinx) -1/2ln(1-sinx)
y'=1/2×1/(1+sinx) ×cosx-1/2×1/(1-sinx) ×(-cosx)
=1/2 ×[cosx/(1+sinx)+cosx/(1-sinx)]
=1/2 ×cosx ×2/(1-sin方x)
=1/cosx
=secx
所以
dy=secxdx